Prepard by :Dr.Bijumon ramalayathil
Contend Editor:M.Nandakumar
Presenter:Rashmi M
Vector Analysis and Geometry
Surface Integrals of vector fields
Objectives
From this unit a learner is expected to achieve the following
- Familiarize with oriented surfaces.
- Learn the definition of surface integral of a vector field.
- Learn the method of finding surface integral of a vector field.
Introduction
In this session we describe surface integrals of vector fields. Recall that in line integrals the orientation of the curve we were integrating along could change the answer. The same thing will hold true with surface integrals. So, before we really get into doing surface integrals of vector fields we first need to introduce the idea of anoriented surface.
Oriented Surfaces
We call a smooth surface S orientable or two-sided if it is possible to define a field of unit vectors on S that varies continuously with position. Any patch or subportion of an orientable surface is orientable. Spheres and other smooth closed surfaces in space (smooth surfaces that enclose solids) are orientable. The surface can be thought of as to have two sidesof which one is arbitrarily chosen positiveside and the other, the negative side. By convention, outer side is chosen as positive. We choose on a closed surface to point outward. Once has been chosen, we say that we have oriented the surface, and call the surface together with its normal field an oriented surface. The vector at any point is called the positive direction at that point.
Example for nonorientable surface
There are nonorientable surfaces also. A well-known example of such a surface is the shown in Fig. 3. When a normal vector, which is given at , is displaced continuously along the curve C in Fig. , the resulting normal vector upon returning to is opposite to the original vector at . A model of a can be made by taking a long rectangular piece of paper, making a half- twist and sticking the shorter sides together so that two points A and the two points B in Fig. coincide.
Remarks
- In the case of a closed surface usually the outer side is taken as the positive side.
- In our discussion that follows, we denote the positive side of the surface by S.
- The unit normal at any point to the surface drawn in the positive side is denoted by n.
- Sometimes a two sided surface S may bounded by a simple closed curve C.
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Surface Integrals of Vector Fields
Suppose S , the positive side of the surface, be divided into a number of elements of area Si. Let P (x, y, z) be an arbitrary point in Si. We take n = n(x, y, z) be the unit normal outward vector drawn at the point P on Si. Then represents the component of F along the normal to S at P.
NotationTo get a compact form hereinafter we denote simply by noting that the value of depends on the point we choose.
Now Si represents the total number of lines of force or total flux over the elemental area Si. Considering flux over all such elemental areas, we form the sum
Then the limit of this sum as each Si tends to zero is defined as the surface integral of F over S, and is denoted by . (This is sometimes called theflux ofFacrossS.)
Hence
where dS=dS(x, y, z) is the vector valued function, called surface area differential, given by
dS = dy dz i + dz dx j + dx dy k.
We note that dS = dS n, where dS is the elementary area of the surface S at the point P and n is the unit normal to the surface at that point. Hence, we have is also denoted by. Hence
Special case If = (x, y, z), then
Remark Another type of surface integral is
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Evaluation of Surface Integrals
Method One(Using Double Integrals) For evaluating surface integral, it is convenient to express them as double integrals taken over orthogonal projection of the surface S on one of the co-ordinate planes. But this is possible only if any line perpendicular to the co-ordinate plane chosen meets the surface S in not more than one point. If the surface does not satisfy this condition, then it may be subdivided into surfaces which satisfy this condition.
Let R be the orthogonal projection of S on the XY plane. Let P (x, y, z) be any point on S. If the acute angle which the outward normal at P (x, y, z) to the surface S makes with the Z-axis, then
cos dS = dx dy,
where dS is the elementary area of the surface S at the point P. Hence
,
where k is the unit vector along Z-axis. Hence
where R is the orthogonal projection of the surface Son the XY plane.
Remark 1
Similarly, if we consider the projection of S on the YZ plane we have
,
where R1 is the projection of S on of S on the YZ plane.
Remark 2
If we consider the projection along the XZ plane we have
,
where R2 is the projection of S on ZX plane.
Method Two Surface integral can also be evaluated using Gauss divergence theorem.Gauss divergence theorem will not be discussed in this session.
Example 1 Evaluate , where F =yzi + zx j + xy k and S is the surface of the sphere x2 +y2 +z2 = 1, which lies in the first octant.
Solution
A vector normal at the point P (x, y, z) on the surface S in the first octant is given by
( x2 +y2+z2 1) = 2xi + 2yj + 2zk.
Thus the unit normal at P (x, y, z) of S is
,
since x2 +y2+z2 = 1 on the surface S.
Now, we note that
where R is the projection of S on the xy plane. Being the projection on the xy plane of the portion of the sphere x2 +y2 +z2 = 4 in the first octant, the region R is bounded by X-axis, Y-axis, and the first quadrant of the circle x2 +y2 = 1. Also z =0 on R.
We find
F . n = (yzi + zx j + xy k) . (xi + yj + zk) = 3 xyz
and n . k = (xi + yj + zk) .k = z.
Since S lies in the first octant, Hence
In polar co-ordinates, points (x, y) in the region R is given by
x = r cos , y = r sin ; 0 r 1, 0 .
Hence,
on simplification.
Example 2 Evaluate , where F =zi x j 3y2z k and S is the surface of the cylinder x2 +y2 = 4 included in the first octant between z = 0 and z = 3.
Solution
A vector normal to x2 +y2 = 4 is given by ( x2 +y2 4) = 2xi + 2yj.
Thus the unit normal at any point (x, y, z) on S is given by
, since x2 +y2 = 4.
Now project S on the ZX plane and call the projection R. (We bring your attention to note that in this case we cannot take projection of S to the xy plane as the surface S is perpendicular to the xy plane.)
Then,
Here = (xy + zx) and = y. Therefore
Example 3 Compute the surface integral of the normal component of F =x2i + yxj + zx k over the triangle with vertices (1,0,0), (0,2,0) and (0, 0, 3).
Solution
The equation of the plane passing through (1,0,0), (0,2,0) and (0, 0, 3) is
fFig 1
i.e.,
i.e., the equation of the plane is 6x + 3y + 2 z = 6 and a vector normal to it is given by
(6x + 3y + 2 z 6) = 6 i + 3 j + 2 k.
Thus the unit normal at any point (x, y, z) on the triangleS is given by
Hence F . n = and n. k =
On S we have z = 3 3x y, so that zx =3 x3 x2yx
Example 4 Evaluate over the surface of the tetrahedron with vertices (1,0,0), (0,2,0) (0, 0, 3), (0, 0, 0), where F =x2i + yxj + zx k.
Solution
We subdivide the surface (tetrahedron) into four surfaces. The triangle with vertices (1,0,0), (0,2,0) and (0, 0, 3); triangle with vertices (1,0,0), (0,2,0) and (0, 0, 3); triangle with vertices (1,0,0), (0,2,0) and (0, 0, 3); and triangle with vertices (1,0,0), (0,2,0) and (0, 0, 3);
(i )We have already computed the integral over one surface in the last example.
(ii)Along the bottom face we have n = k and hence F. n = zx, but since z= 0 the integral over the bottom face is 0.
(iii)On the face at the left we have n = j and hence F. n = yx which is also 0 since y = 0 there.
(iv)On the rear face, in the yz plane, n = i and F. n = x2 = 0.
It follows that there is only one non zero contribution to the surface integral and hence
Example 5 Evaluate , where F = (x+ y)i + x j + z k and S is the surface of the cube bounded by the planes x = 0, x = 1, y = 0, y = 1, z = 0, z = 1.
Solution
Since cube has 6 faces, the given surface of the cube bounded by the planes x = 0, x = 1, y = 0, y = 1, z = 0, z = 1 can be considered as the union of the six surfaces (faces) viz., S1, S2, S3, S4, S5 and S6 whose equations are x = 0, x = 1, y = 0, y = 1, z = 0 and z = 1 respectively. The normal vectors to each surface (face) is given by n1, n2, n3, n4, n5 and n6 etc.
Now on S1 : n1 = i; x = 0, F = yi + z k and y varies from 0 to 1 and z varies from 0 to 1.
On S2 : n2 = i ; x = 1, F = (1 + y)i + j + z k and y varies from 0 to 1 and z varies from 0 to 1.
On S3 : n3 = j ; y = 0, F = xi + x j + z k and x varies from 0 to 1 and z varies from 0 to 1.
On S4 : n4 = j ; y = 1, F = (x + 1)i + x j + z k and x varies from 0 to 1 and z varies from 0 to 1.
On S5 : n5 = k ; z = 0, F = (x + y)i + x j and x varies from 0 to 1 and y varies from 0 to 1.
On S6 : n6 = k ; z = 1, F = (x + y)i + x j + k and x varies from 0 to 1 and y varies from 0 to 1.
Adding all the above, we have
Summary
In this session we have defined surface integrals of vector fields.We have also discussed the idea of oriented surfaces. The method of finding surface integrals of vector fields have been described in this session. Before concluding the session try the following assignments.
Assignments
1.Evaluate, where F =xi + y j 2 z k and S is the surface of the sphere x2 + y2 + z2 = a2 above the xy plane.
2.Evaluate, where F =zi + xj y2z k and S is the surface of the cylinder x2 + y2 = 1 included in the first octant between planes z = 0 and z = 2.
3. Evaluate, where F = 4xi 2 y2 j + z2k and S is the region bounded by x2 + y2 = 4, z = 0, z = 3.
4.Evaluate, where F = (x + y2)i - 2 xj + 2 yz k and S is the surface of the plane 2x + y +2z = 6 in the first octant.
5.Evaluate, where F =xyi x2 j + (x + z)k and S is that region of the plane 2x + 2y+ z = 6 bounded in the first octant.
6.Evaluate, where F =yi + 2x j z k and S is the surface of the plane 2x + y = 6 in the first octant cut off by the plane z = 4.
7.Evaluate, where F = 4xzi y2 j + yzk and S is the surface bounded by the cube x = 0, y = 0, z = 0; x = 1, y = 1, z = 1.
8. Show that where S is a sphere of radius r and center at origin and F = x3i + y3j + z3 k.
9. Evaluate , where S is a sphere of radius 1 and center at origin.
10.Let F = xy2 i + yz2 j + zx2 k . Evaluate over the sphere given by x2 + y2 + z2 =1.
FAQ
1. How many normal vectors are there for every point in a two sided surface.
Ans. A two sided surface has a tangent plane at every point (except possibly along the boundary). Making this assumption means that every point will have two unit normal vectors, and this means that every surface will have two sets of normal vectors. The set that we choose will give the surface an orientation.
2. A two sided surface S is always bounded by a simple closed curve C. True/False. Comment.
Ans. There are two sided surfaces not bounded by a simple closed curve. For there are closed surfaces, that are boundary of some solid region. A good example of a closed surface is the surface of a sphere.
4. What can you say about the orientation of a closed surface.
Ans. We say that the closed surfaceShas apositiveorientation if we choose the set of unit normal vectors that point outward from the solid regionE that is enclosed by S while thenegativeorientation will be the set of unit normal vectors that point in towards the regionE. Note that this convention is only used for closed surfaces.
5. In the evaluation of , where F =zi x j 3y2z k and S is the surface of the cylinder x2 +y2 = 4 included in the first octant between z = 0 and z = 3, is it possible to consider projection on XY plane. Why?
Ans. No. For evaluating surface integral, it is convenient to express them as double integrals taken over orthogonal projection of the surface S on one of the co-ordinate planes. But this is possible only if any line perpendicular to the co-ordinate plane chosen meets the surface S in not more than one point. In the evaluation of the given line integral it is not possible to consider projection on XY plane since some lines perpendicular to the XY meets the given surface at more than one point.
Quiz
If is a continuous vector field defined on an oriented surface S with unit normal then the surface integral of over S is given by
(a)
(b)
(c)
(d) None of the above.
Ans. (a
GLOSSARY
Orientable surface:We call a smooth surface S orientable or two-sided if it is possible to define a field of unit vectors on S that varies continuously with position
surface area differential:
dS = dy dz i + dz dx j + dx dy k.
REFERENCE
- Murray R. Spiegel, Vector Analysis, Schaum Publishing Company, New York.
- N. Saran and S. N. Nigam, Introduction to Vector Analysis, Pothishala Pvt. Ltd., Allahabad.
- Erwin Kreyszig, Advanced Engineering Mathematics, John Wiley & Sons, 1999.
- Shanti Narayan, A Text Book of Vector Calculus, S. Chand & Co., New Delhi.
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